Let a1 = 1 and for n ≥ 1, an+1 = 1/2 an + (n²-2n-1)/(n²(n+1)²). Then Σ (n=1 to ∞) (an - 2/n²) is equ

Q: Why did we choose $b_n = a_n - \frac{2}{n^2}$? How would I know to do that in an exam?

This is a classic pattern recognition skill in JEE Advanced. Whenever you see a recurrence relation $a_{n+1} = f(a_n) + g(n)$ and you are asked to find the sum of a specific expression involving $a_n$, that expression is almost always the key to a transformation. By defining $b_n$ as that expression, you are essentially 'shifting' the sequence to a simpler form where the non-homogeneous part $g(n)$ is absorbed into the definition of $b_n$.

Q: What if the numerator didn't simplify to zero? Does this method always work?

If the numerator didn't simplify to zero, it would mean the substitution was either incorrect or the problem was designed differently. In competitive exams, these problems are meticulously crafted. If you find yourself with a messy remainder, pause and re-check your algebraic expansion. The 'magic' cancellation is a hallmark of these problems, confirming you are on the right track.

Q: Can we use the sum of an infinite G.P. even if the terms are negative?

Absolutely! The formula $S = \frac{a}{1-r}$ works for any geometric series as long as the common ratio $|r| < 1$. The sign of the first term $a$ simply dictates the sign of the sum. Here, $b_1 = -1$ and $r = \frac{1}{2}$, so the sum is negative, which is perfectly valid.

Question

Let $a_1 = 1$ and for $n \ge 1, a_{n+1} = \frac{1}{2} a_n + \frac{n^2-2n-1}{n^2(n+1)^2}$. Then $\sum_{n=1}^\infty (a_n - \frac{2}{n^2})$ is equal to ......... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the base of a mountain. You look up, and the path ahead seems shrouded in fog—a complex recurrence relation:

$$a_{n+1} = \frac{1}{2} a_n + \frac{n^2-2n-1}{n^2(n+1)^2}$$

Many students would immediately try to find the general term $a_n$ by iteration, only to get lost in a labyrinth of partial fractions and infinite sums. But in the world of JEE Advanced, we don't climb the mountain by brute force; we look for the hidden path.

The target sum, $\sum_{n=1}^{\infty} (a_n - \frac{2}{n^2})$, is your lighthouse. It is telling you exactly where to go.

### The Elegant Transformation

Whenever you see a recurrence relation paired with a specific expression in a summation, that expression is your golden ticket. Let us define a new sequence, $b_n$, such that:

$$b_n = a_n - \frac{2}{n^2}$$

By doing this, we are not just simplifying the notation; we are changing our perspective. We are shifting our focus from the messy sequence $a_n$ to a cleaner, more well-behaved sequence $b_n$.

This implies that $a_n = b_n + \frac{2}{n^2}$, and consequently, $a_{n+1} = b_{n+1} + \frac{2}{(n+1)^2}$. Now, watch as we substitute these into our original recurrence relation:

$$b_{n+1} + \frac{2}{(n+1)^2} = \frac{1}{2} \left( b_n + \frac{2}{n^2} \right) + \frac{n^2-2n-1}{n^2(n+1)^2}$$

### The Magic of Cancellation

This is the moment where the tension breaks. As we distribute the $\frac{1}{2}$ and isolate $b_{n+1}$, we are left with a collection of fractional terms on the right side:

$$b_{n+1} = \frac{1}{2} b_n + \frac{1}{n^2} - \frac{2}{(n+1)^2} + \frac{n^2-2n-1}{n^2(n+1)^2}$$

When we find a common denominator, $n^2(n+1)^2$, and combine the numerators, something beautiful happens. The numerator expands to:

$$(n+1)^2 - 2n^2 + (n^2 - 2n - 1)$$

Expanding this gives $(n^2 + 2n + 1) - 2n^2 + n^2 - 2n - 1$. Look closely—the $n^2$ terms cancel, the $n$ terms cancel, and the constants cancel. Everything vanishes!

We are left with $b_{n+1} = \frac{1}{2} b_n$. The mountain has flattened into a simple, beautiful **Geometric Progression**.

### The Final Ascent

Now that we have identified our sequence as a G.P. with a common ratio $r = \frac{1}{2}$, we only need the starting point. Using our substitution $b_n = a_n - \frac{2}{n^2}$ for $n=1$, we find:

$$b_1 = a_1 - \frac{2}{1^2} = 1 - 2 = -1$$

The sum of an infinite G.P. is given by $S = \frac{b_1}{1-r}$. Substituting our values, we get:

$$S = \frac{-1}{1 - 1/2} = \frac{-1}{1/2} = -2$$

We have reached the summit. The seemingly impossible problem has been solved with elegance and precision. The final answer is **-2**.

Animated Solution for Mathematics - Sequence and Series: Let $a_{1} = 1$ and for $n \geq 1, a_{n + 1} = \frac{1}{2} a_{n} + \frac{n ^{2} - 2 n - 1}{n ^{2} ( n + 1 ) ^{2}}$ . Then $\sum_{n = 1}^{\infty} (a_{n} - \frac{2}{n ^{2}})$ is equal to ......... .

Enter Numerical Value:

Visualized Solution

Objective: $\sum_{n = 1}^{\infty} (a_{n} - \frac{2}{n ^{2}})$

Substitution: $b_{n} = a_{n} - \frac{2}{n ^{2}}$

Substitute into Recurrence

Isolate $b_{n + 1}$

Simplify the Fractions

Numerator Expansion

Result: $b_{n + 1} = \frac{1}{2} b_{n}$

Calculate $b_{1}$

Sum of Infinite G.P.

Final Answer: $- 2$

Key Takeaways

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

The Elegant Transformation

The Magic of Cancellation

The Final Ascent

Similar Questions

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Let $a_{n}$ be the $n^{t h}$ term of a G.P. of positive terms. If $\sum_{n = 1}^{100} a_{2 n + 1} = 200$ and $\sum_{n = 1}^{100} a_{2 n} = 100$ , then $\sum_{n = 1}^{200} a_{n}$ is equal to:

If the sum of an infinite $GP$ $a, a r, a r^{2}, a r^{3}, \dots$ is $15$ and the sum of the squares of its each term is $150$ , then the sum of $a r^{2}, a r^{4}, a r^{6}, \dots$ is :

If $A = \sum_{n = 1}^{\infty} \frac{1}{( 3 + ( - 1 ) ^{n} ) ^{n}}$ and $B = \sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n}}{( 3 + ( - 1 ) ^{n} ) ^{n}}$ , then $\frac{A}{B}$ is equal to:

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If each term of a geometric progression $a_{1}, a_{2}, a_{3}, \dots$ with $a_{1} = \frac{1}{8}$ and $a_{2} \neq = a_{1}$ , is the arithmetic mean of the next two terms and $S_{n} = a_{1} + a_{2} + \dots + a_{n}$ , then $S_{20} - S_{18}$ is equal to

Let $A_{1}, A_{2}, A_{3}, \dots$ be squares such that for each $n \geq 1$ , the length of the side of $A_{n}$ equals the length of diagonal of $A_{n + 1}$ . If the length of $A_{1}$ is 12 cm, then the smallest value of $n$ for which area of $A_{n}$ is less than one, is

Enter Numerical Value:

Visualized Solution

Objective: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑n=1∞​(an​−n22​)

Substitution: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bn​=an​−n22​

Substitute into Recurrence

Isolate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bn+1​